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Heat The big ideas!

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**Phases of Matter Solid – lattice work, has shape**

Liquid – fluid, takes shape of container Gas – fluid, needs sealed container (Plasma) – most energetic

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**Temperature Temperature – measure of hotness**

degrees Celsius or degrees Centigrade or °C Fahrenheit scale Kelvin Scale (doesn’t use the symbol) Absolute zero Freezing of water Boiling of water -273 °C 0 °C 100 °C -460 °F 32 °F 212 °F 0 K 273 K 373 K

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**Temperature and Molecular Motion**

The higher the temperature, the faster the molecules are moving. The faster the molecules are moving, the more energy they have For the same molecule: Gas > Liquid > Solid Gas has more energy than a solid

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**Heat Heat is the flow of thermal energy**

from hot substance to cold substance related to the amount of thermal energy both temperature and mass are important Objects in contact tend to reach same temperature thermal equilibrium

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**Heat Measure of energy transferred**

calorie – cal (little c) Calorie = 1000 calories = kilocalorie (big C) Joule (1 calorie = J) BTU (British Thermal Unit) Transfer of Heat requires temperature difference Heat moves from high to low temperature Thermal expansion most materials get bigger as they get hotter

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**Specific Heat Capacity**

symbol “c” Unit: Energy/mass * temperature difference J/kg °C or cal/g °C or BTU/ lb °F measures effect of heat on change in temperature c of water is 1 cal/g °C this is high c of metals is lower – easier to change temperature Q = m c T

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**Thermal Equilibrium Problem**

A 2.0 kg block of aluminum at 60 C is placed in contact with a 3.0 kg block of steel at 30 C. If no heat is lost to the environment, what will be the final temperature of the blocks? cal = 900 J/kg C csteel = 470 J/kg C

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**Thermal Equilibrium Problem**

The aluminum block will lose heat to the steel block The heat lost by the aluminum equals that gained by the steel The final temperature will be some place between the initial temperatures of the blocks.

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**Qal + Qsteel = 0 (aluminum loses while steel gains)**

mal cal Tal + mst cst Tst = 0 mal cal (Tf-Ti)al + mst cst (Tf-Ti)st = 0 2.0(900)(Tf-60) + 3.0(470)(Tf-30) =0 1,800Tf – 108, ,410Tf – 42,300 = 0 3,210Tf – 150,300 = 0 Tf = 150,300/3,210 = 46.8 C

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**Thermal Equilibrium Problem 2**

A 9.0 kg block of aluminum at 60 C is placed in contact with a 1.0 kg block of steel at 30 C. If no heat is lost to the environment, what will be the final temperature of the blocks? cal = 900 J/kg C csteel = 470 J/kg C

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Problem #2 Qal + Qsteel = 0 (aluminum loses while steel gains) mal cal Tal + mst cst Tst = 0 mal cal (Tf-Ti)al + mst cst (Tf-Ti)st = 0 9.0(900)(Tf-60) + 1.0(470)(Tf-30) =0 8100Tf – 486, Tf – 14,100 = 0 8570Tf – = 0 Tf = 500,100/8,570 = 58.3 C

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**Heat Transfer Conduction Convection Radiation**

Heat transfers along an object metals have high conduction Convection heat transfers when a fluid moves hot air rises and cold air sinks creates wind and “weather” Radiation any object with thermal energy emits radiation electromagnetic wave black absorbs radiant energy better than white

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**Changing Phase Gas Liquid Solid Energy evaporation condensation**

add heat condensation remove heat Heat of vaporization 540 cal/g for water 2257 kJ/kg Liquid melting add heat Heat of fusion 80 cal/g for water 334 kJ/kg freezing remove heat Solid

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**Freezing and Melting Freezing Melting temperature remains constant**

from liquid to solid loses energy (heat) to freeze a liquid Melting From solid to liquid requires heat to melt solid temperature remains constant 80 cal/g (334 kJ/kg) of heat lost to freeze water Called Heat of fusion

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**Evaporation and Condensation**

Heat required to change liquid to gas boiling Condensation Heat released when gas change to liquid temperature of solid and liquid are the same 540 cal/g (2257 kJ/kg) to evaporated water Called Heat of vaporization

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540 cal 80 cal

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**How much heat is required to raise the temperature of 3 g of water from 23 °C to 39 ° C?**

How much heat is required to melt 3 g of ice at 0 ° C? How much heat is required to vaporize 4 g of water to steam at 100 ° C? How much heat is required to take 2 g of ice from 0 ° C to steam at 100 ° C?

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**How much heat is required to raise the temperature of 3 g of water from 23 °C to 39 °C?**

Q=mc(Tf-Ti)= 3g x 1 cal/g °C x 16 °C = 48 cal How much heat is required to melt 3 g of ice at 0 °C? 3 g x 80 cal/g = 240 cal How much heat is required to vaporize 4 g of water to steam at 100 °C? 4 g x 540 cal/g = 2160 cal How much heat is required to take 2 g of ice from 0 °C to steam at 100 °C? Ice to water g x 80 cal/g = 160 cal 0 °C to 100 °C 2g x 1 cal/g °C x 100 °C = 200 cal Water to steam g x 540 cal/g = cal total cal

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Heat Chapter 9.

Heat Chapter 9.

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